math
posted by isa .
suppose an arithmetic sequence and a geometric sequence with common ration r have the same first two terms. show that the third term of the geometric series is r^2/(2r1) tomes the third term of the arithmetic sequence

the first 3 terms of the AS are a, a+d, a+2d
the first 3 terms of the GS are a, ar, ar^2
you said the second terms are equal
then a+d = ar
d = ara
so third term of GS/third term of AS
= ar^2/(a+2d)
= ar^2(a + 2(ara)
= ar^2/(a + 2ar  2a)
= ar^2/(2ar  a) now divide top and bottom by a
= r^2/(2r1) 
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